Q2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Solution

Let the equilateral triangle have side length \(2a\). Since the base lies along the \(y\)-axis and its midpoint is at the origin, the endpoints of the base must be equally spaced above and below the origin.

Thus, the coordinates of the base vertices are \[ A(0,a) \quad \text{and} \quad B(0,-a) \]

The length of the base is \[ \begin{aligned} AB &= \sqrt{(0-0)^2 + (a - (-a))^2} \\&= \sqrt{(2a)^2} \\&= 2a \end{aligned} \] which matches the given side length.

In an equilateral triangle, the perpendicular from the opposite vertex to the base passes through the midpoint. Since the midpoint is at the origin, the third vertex lies on the \(x\)-axis. The height \(h\) of an equilateral triangle of side \(2a\) is

\[ \begin{aligned} h &= \sqrt{(2a)^2 - a^2} \\ &= \sqrt{4a^2 - a^2} \\ &= \sqrt{3a^2} = \sqrt{3}\,a \end{aligned} \]

Therefore, the third vertex can be located on either side of the \(y\)-axis as \[ C(\sqrt{3}\,a, 0) \quad \text{or} \quad C(-\sqrt{3}\,a, 0) \]

Hence, the vertices of the equilateral triangle are \[ (0,a),\ (0,-a),\ (\sqrt{3}a,0) \] or \[ (0,a),\ (0,-a),\ (-\sqrt{3}a,0) \]

Q3. Find the distance between \(P (x_1,\; y_1)\) and \(Q (x_2,\; y_2)\) when :
(i) PQ is parallel to the y-axis,
(ii) PQ is parallel to the x-axis.

Solution

Let \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) be two points in the Cartesian plane. The distance between two points is generally found using the distance formula, but here we simplify it based on the direction of the line segment \(PQ\).

When \(PQ\) is parallel to the \(y\)-axis, both points have the same \(x\)-coordinate, that is \(x_1 = x_2\). The distance between the points is the vertical difference in their \(y\)-coordinates:

\[ \begin{aligned} PQ &= |y_2 - y_1| \end{aligned} \]

When \(PQ\) is parallel to the \(x\)-axis, both points have the same \(y\)-coordinate, that is \(y_1 = y_2\). The distance between the points is the horizontal difference in their \(x\)-coordinates:

\[ \begin{aligned} PQ &= |x_2 - x_1| \end{aligned} \]

Thus, if a line segment is parallel to the \(y\)-axis, its length depends only on the difference in \(y\)-coordinates, and if it is parallel to the \(x\)-axis, its length depends only on the difference in \(x\)-coordinates.

Q4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Solution

Let the required point on the x-axis be \((x, 0)\). Since this point is equidistant from \((7, 6)\) and \((3, 4)\), the distances from these two points must be equal.

\[ \begin{aligned} \sqrt{(7 - x)^2 + (6)^2} &= \sqrt{(3 - x)^2 + (4)^2} \end{aligned} \]

Squaring both sides to remove the square roots, we get

\[ \begin{aligned} (7 - x)^2 + 36 &= (3 - x)^2 + 16 \end{aligned} \]

Expanding both sides,

\[ \begin{aligned} 49 - 14x + x^2 + 36 &= 9 - 6x + x^2 + 16 \end{aligned} \]

Simplifying,

\[ \begin{aligned} 85 - 14x &= 25 - 6x \\ -14x + 6x &= 25 - 85 \\ -8x &= -60 \\ x &= \frac{60}{8} = \frac{15}{2} \end{aligned} \]

Therefore, the required point on the x-axis is \(\left(\dfrac{15}{2}, 0\right)\).

Q5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).

Solution

Let the coordinates of the midpoint of the line segment \(PB\) be \((x_1, y_1)\). Using the midpoint formula, we find the coordinates of the midpoint of \(P(0, -4)\) and \(B(8, 0)\).

\[ \begin{aligned} x_{1} &= \frac{0 + 8}{2} = 4 \\ y_{1} &= \frac{-4 + 0}{2} = -2 \end{aligned} \]

Thus, the midpoint is \((4, -2)\). Now, we find the slope of the line passing through the origin \((0, 0)\) and the point \((4, -2)\).

\[ \begin{aligned} m &= \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ &= \frac{0 - (-2)}{0 - 4} \\ &= \frac{2}{-4} \\ &= -\frac{1}{2} \end{aligned} \]

Therefore, the slope of the required line is \(-\dfrac{1}{2}\).

Q6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.

Solution

Let the given points be \(A(4,4)\), \(B(3,5)\), and \(C(-1,-1)\). To show that these points form a right angled triangle without using the Pythagoras theorem, we verify whether two sides are perpendicular by comparing their slopes.

Slope of line \(AB\) is

\[ \begin{aligned} m_{AB} &= \frac{5 - 4}{3 - 4} \\ &= \frac{1}{-1} \\ &= -1 \end{aligned} \]

Slope of line \(BC\) is

\[ \begin{aligned} m_{BC} &= \frac{-1 - 5}{-1 - 3} \\ &= \frac{-6}{-4} \\ &= \frac{3}{2} \end{aligned} \]

Slope of line \(AC\) is

\[ \begin{aligned} m_{AC} &= \frac{-1 - 4}{-1 - 4} \\ &= \frac{-5}{-5} \\ &= 1 \end{aligned} \]

Since the product of slopes of lines \(AB\) and \(AC\) is

\[ \begin{aligned} m_{AB} \times m_{AC} &= (-1)(1) = -1 \end{aligned} \]

this shows that lines \(AB\) and \(AC\) are perpendicular. Therefore, the angle at point \(A(4,4)\) is a right angle.

Hence, the points \((4,4)\), \((3,5)\), and \((-1,-1)\) are the vertices of a right angled triangle.

Q7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

Solution

The line makes an angle of \(30^\circ\) with the positive direction of the \(y\)-axis measured anticlockwise. Since the positive \(y\)-axis is perpendicular to the positive \(x\)-axis, the angle the line makes with the positive \(x\)-axis is

\[ \begin{aligned} \theta &= 90^\circ + 30^\circ = 120^\circ \end{aligned} \]

The slope of a line making an angle \(\theta\) with the positive \(x\)-axis is given by \(m = \tan \theta\). Therefore, the slope of the given line is

\[ \begin{aligned} m &= \tan(120^\circ) \\ &= \tan(180^\circ - 60^\circ) \\ &= -\tan(60^\circ) \\ &= -\sqrt{3} \end{aligned} \]

Hence, the slope of the line is \(-\sqrt{3}\)

Q8. Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.

Solution

Let the given points be \(A(-2,-1)\), \(B(4,0)\), \(C(3,3)\), and \(D(-3,2)\). To prove that these points form a parallelogram without using the distance formula, we show that the diagonals bisect each other by comparing their midpoints.

Midpoint of diagonal \(AC\) is

\[ \begin{aligned} M_1 &= \left( \frac{-2 + 3}{2}, \frac{-1 + 3}{2} \right) \\ &= \left( \frac{1}{2}, 1 \right) \end{aligned} \]

Midpoint of diagonal \(BD\) is

\[ \begin{aligned} M_2 &= \left( \frac{4 + (-3)}{2}, \frac{0 + 2}{2} \right) \\ &= \left( \frac{1}{2}, 1 \right) \end{aligned} \]

Since the midpoints of diagonals \(AC\) and \(BD\) are the same, the diagonals bisect each other.

Therefore, the points \((-2,-1)\), \((4,0)\), \((3,3)\), and \((-3,2)\) are the vertices of a parallelogram.

Q9. Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2).

Solution

Let the given points be \(A(3,-1)\) and \(B(4,-2)\). The angle between the line joining these points and the positive \(x\)-axis can be found using the slope of the line.

First, we calculate the slope \(m\) of the line \(AB\):

\[ \begin{aligned} m &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{-2 - (-1)}{4 - 3} \\ &= \frac{-1}{1} \\ &= -1 \end{aligned} \]

If \(\theta\) is the angle that the line makes with the positive direction of the \(x\)-axis, then

\[ \begin{aligned} \tan \theta &= m = -1 \end{aligned} \]

Thus, the angle of inclination is

\[ \begin{aligned} \theta &= \tan^{-1}(-1) = -45^\circ \end{aligned} \]

Since the angle is measured anticlockwise from the positive \(x\)-axis, the corresponding positive angle is

\[ \begin{aligned} \theta &= 180^\circ - 45^\circ = 135^\circ \end{aligned} \]

Therefore, the angle between the \(x\)-axis and the line joining the given points is \(135^\circ\)

Q10. The slope of a line is double of the slope of another line. If tangent of the angle between them is \(\frac{1}{3}\), find the slopes of the lines.

Solution

Let the slope of the first line be \(m\). Since the slope of the second line is double of the first, its slope is \(2m\). The tangent of the angle between two lines with slopes \(m_1\) and \(m_2\) is given by

\[ \begin{aligned} \tan \theta = \frac{m_2 - m_1}{1 + m_1 m_2} \end{aligned} \]

Substituting \(m_1 = m\) and \(m_2 = 2m\), we get

\[ \begin{aligned} \tan \theta &= \frac{2m - m}{1 + m \cdot 2m} \\ &= \frac{m}{1 + 2m^2} \end{aligned} \]

Given that \(\tan \theta = \frac{1}{3}\), we have

\[ \begin{aligned} \frac{1}{3} &= \frac{m}{1 + 2m^2} \\ 1 + 2m^2 &= 3m \\ 2m^2 - 3m + 1 &= 0 \\ 2m^2 - 2m - m + 1 &= 0 \\ 2m(m - 1) - 1(m - 1) &= 0 \\ (m - 1)(2m - 1) &= 0 \\ m - 1 &= 0 \\ m &= 1 \\ 2m - 1 &= 0 \\ m &= \frac{1}{2} \end{aligned} \]

Hence, the possible slopes of the lines are \(m = 1\) and \(m = \frac{1}{2}\), and the corresponding second slopes are \(2\) and \(1\) respectively

Q11. A line passes through \((x_1, y_1)\) and \((h, k)\). If slope of the line is m, show that \(k – y_1 = m (h – x_1).\)

Solution

Let a line pass through the points \((x_1, y_1)\) and \((h, k)\), and let its slope be \(m\). By definition, the slope of a line joining two points is the ratio of the change in \(y\)-coordinates to the change in \(x\)-coordinates.

Thus, the slope \(m\) of the line is given by

\[ \begin{aligned} m &= \frac{k - y_1}{h - x_1} \end{aligned} \]

Multiplying both sides by \((h - x_1)\), we obtain

\[ \begin{aligned} k - y_1 &= m (h - x_1) \end{aligned} \]

Hence, it is proved that \(k - y_1 = m (h - x_1)\)